Computer Science, asked by vipulvarsha7267, 1 year ago

JIT University offering degree courses to students has decided to provide scholarship based on the following details: Branch of study Score (%) Scholarship % Remarks Arts Score is at least 90 50 The student is eligible only for one scholarship% even if both the score conditions are valid for the given branch of study. In such cases, students are eligible for the highest scholarship% applicable among the two. Arts Score is an odd number 5 Engineering Score is more than 85 50 Engineering Score is divisible by 7 5 If there are 500 students who have joined the university, write a pseudo-code to calculate and display the final fees to be paid by each student. You may accept the branch of study, score and course fee as inputs for each student and calculate the final fees to be paid by each student based on formulae given below: Scholarship amount=course fee * (scholarship%) Final fee= course fee - scholarship amount

Answers

Answered by singhrishugpj
11

Answer:

for(cont=1,cont<=500,cont=cont+1;) //for 500 student//

scholarship=0

input Branch_Of_study,score,course_fee

if(Branch_Of_study=="Arts")then

if(Score>=90)then

scholarship=50

else if(score%2!= 0)then

scholarship=5

end if                              //*closing of  score condition*//

else if(Branch_Of_stud=="Engineering ")then

if(score>85)then

scholarship=50

else if(score%7==0)then

scholarship=5

end if                 //*closing  of score condition*//

end if               //*closing of branch condition*//

scholarship_Amount=course_fee*scholarship/100

Final_Fee=Course_fee-scholarship_Amount

display Final_fee

end for         //*closing of loop*//

Explanation:

step 1: create a loop for 500 student

step 2: assign scholarship=0 for every time ending of one student fee scholarship will again become zero

step 3: check the condition for branch

step 4: then check the condition for score

step 5: and  then assign the scholarship base on condition

step 6: then calculate the amount of scholarship

again repeat the loop for next and so on........  

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